The mean free path of molecules in an ideal gas A is half that of another ideal gas B. The diameter of the spherical molecules of gas A is twice the diameter of the molecules of B. If the number densities of gases A and B are n_A and n_B respectively, the correct option is:

Solution & Explanation

Governing formula: the mean free path of a gas molecule is λ=1/(√2·π·d²·n), where d is the molecular diameter and n the number density. Step 1 — Write the ratio λ_A/λ_B (the √2π cancels). λ_A/λ_B=(d_B²·n_B)/(d_A²·n_A). Step 2 — Put in the data: λ_A=(1/2)λ_B, so λ_A/λ_B=1/2, and d_A=2d_B. 1/2=(d_B²·n_B)/((2d_B)²·n_A)=(d_B²·n_B)/(4d_B²·n_A)=n_B/(4n_A). Step 3 — Solve for n_A. 1/2=n_B/(4n_A) → 4n_A=2n_B → n_A=(1/2)n_B. This matches option D: n_A=(1/2)n_B. Why the answer is sensible: A has bigger molecules (d² is 4×) which alone would shorten λ a lot; to make λ_A only half of λ_B (not a quarter), A must be LESS dense. The d² effect and the chosen λ ratio together give exactly n_A=½n_B.